# Mathematics Grade 12 LP March Test 2026 Paper & Memo

> Grade 12 Mathematics question paper and memo from the LP March Test 2026, 19 pages. Read online or get the free PDF download.

- Page: https://www.hlayiso.com/library/grade-12-mathematics-2026-lp-march-test-question-paper-and-memorandum/b412e4b4-7a4a-4797-be6b-d30db279b5dc
- Updated: 2026-09-28
- Subject: Mathematics
- Grade: Grade 12
- Document type: Question paper and memo
- Year: 2026
- Exam period: LP March Test
- Curriculum: CAPS
- Pages: 19

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## About this document

Mathematics · Grade 12 · LP March Test · 2026. Question paper and memorandum, 19 pages. Read online or download the PDF.

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## Document text

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NSC

REPUBLIC OF ROUTH AFRICA

PERE: I_IM POPC}
PROVINCIAL GOVERNMENT

DOPPARTMENT Or

EDUCATION

MARKS: 100 DURATION: 2 HOURS

This Question papers consists of 8 pages and 1 information sheet.

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INSTRUCTIONS AND INFORMATION

1. This question paper consists of EIGHT questions.
2. Clearly show ALL calculations, diagrams, et cetera which you have used in

determining the answers.

3. Number the answers correctly accord un g system used in the

question paper.

5. You may use an approved scientific n- grammable and non-

graphical), unless stated otherwise.
6. Ifnecessary, round off answers to TWO decimal places, unless stated otherwise.
7. Diagrams are NOT necessarily drawn to scale.

8. Write neatly and legibly.

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(3)
(4)
[7]
QUESTION 2
2.1 Given the finite arithmetic sequi 3-87
2.1.1 Write down the fourth te m (dd)
2.1.2. Calculate the number of t (3)
2.1.3. Calculate the sum of all the negative numbers in the sequence. (3)
2.2 The tenth and the seventeenth terms of an arithmetic sequence are
21 and 49 respectively
2.2.1 Determine the common difference of the sequence. (3)
(3)
23
(1)
(4)
[18]

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NSC

quadratic number pattern.
two terms of the number pattern.

erm of the number pattern in the

3.2.1 Determine the n‘” term

3.2.2 Calculate the sum of the fit

3.3 The first and second term of a geom eries is
‘tai re) 1OS,CG.
and (4x? — 16) respectively. De eiinenaneealiets

the series will converge.

en as (2x — 4)
of x for which

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(1)

(4)

(1)
(2)

(4)
[12]

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NSC

QUESTION 4

Sketched below are the graphs of f(x) = —2x? + 4x +16 and g(x) = 2x + 4.
A and B are the x-intercepts of f C is the turning point of f

4.1 Calculate the coordinates of A and B. (3)
4.2 Determine the coordinates of C, the turning points of /- (2)
43 Write down the range of f. ()
44 The graph of h(x) = f(x + p) + q has a maximum value of 15 at x = 2.
Determine the values of p and qg (3)
4.5 Determine the equation of g~', the inverse of g, in the form y =... (2)
4.6 For which value(s) of x will g7().g() = 0? (2)
[13]

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QUESTION 5

Given the exponential function: f(x) = 3*

sit Write down the range of f- (dd)
5.2 Determine the equation of f~! in the form y=... (2)
5:3 The point M(a; 2) lies on f~?. Calculate the value of a. (2)
5.4 Sketch, in your answer book, the g f,}, showing clearly ALL
intercepts with axes (include t [ ach graph). Draw and
name the line of symmetry be (5)
5:5 Is f~! a function? Justify your ans (2)
[12]
QUESTION 6
6.1 Ifcos@ = -< where 180° < 6 < 360°, determine, without using a
calculator, the value of:
6.1.1 sin? 6 (3)
6.1.2 cos (6 — 135) (4)
6.2 Prove that 2cos?(45° + 1) = a nee (4)
6.3 Consider the expressi
6.3.1 Prove that sin(A - (2)
6.3.2 Simplify the followi ingl 1: sin4dx —sinl0x. (2)
6.3.3. Hence, determine the. solutioi Ox = sin3x for
x € (0°; 30°] ae (5)
[20]
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QUESTION 7

7.1 Write down:

7.1.1 The period of f dd)
7.1.2 The amplitude of g (1)

7.2. Without using a calculator, determine the value of x for which
cos2x = —sinx in the interval x € [—180° ; 180°]. (6)

7.3 Use the graphs above to determine how many degrees apart are points A and B
from each other? (2)
[10]

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NS!

QUESTION 8

, A, B and C are points in the same horizontal plane. D is a point directly
that is DC 1 AC and DC 1 BC. It is given that ACB = 100°, CAD = 30°,
10 units and BC = 8 units.

8.1 Calculate the length of:

8.1.1 AC (2)

8.1.2 AB (3)

8.2 If it is further given that ABD = 73,4°, calculate the size of ADB. (3)
[8]

TOTAL: [100]

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INFORMATION SHEET
Pr. =btlb? —4ac
2a
A= P(l+ni) A=P(I-ni) A=(I-i! A=P(I+i)’
T= a4(n-l)d S,=2[2a+(n-Dd]
T, =ar™" S, ae), r#l
r=

petleton!

soe tim LF

d=, -x) +(¥, -yy

ysme+e Y-y, = m(x-%)

(x-a)? +(y-bY =r?

In AABC: Be nO
sinA sinB sinc

a? =b? +c? —2be.cosA
area AABC = Fabsinc

sin(a+ f) =sinacos B+cosasinB

cos(a + 8) =cosa cos S—sinasin 8

cos? a@—sin® @

cos2a= 1-2sin? a
2cos? a@—1
gue
n
n(A)
P(A) =——
8)
poathk

sin(a— 8)= sina oos B—cosasinB

cos(a—£8)=cosacos §+sinasin B

sin2@=2sinacosa

Yo,-3F
214

n

o

P(A or B) = P(A) + P(B)—P(A and B)

pe LE DO-P)
De-xy

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PROVINCIAL GOVERNMENT

REPUBLIC OF ROUTH AFRICA

PEPARTMENT Or

EDUCATION

MARKS: 100

These marking guidelines consists of 11 pages

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NOTE:

e Ifa candidate answers a question TWICE, only mark the FIRST attempt.

e  Ifacandidate crossed out an attempt of a question and not redone the question,
mark the crossed-out version.

© Consistent accuracy applies in ALL aspects of the marking guidelines. Stop marking
at the second calculation error.

«~~ Assuming answers/values to solve a problem is NOT acceptable.

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Question 1
L.A} x?-— 2x +1>0
@=D(x-1) >0 Y factors

Critical value: x =1
x<1 or x>1

¥ ¥ correct notation

Sy2 = -990

(Accuracy)
OR
sxERx#l1
¥VxERxF1
(Accuracy)
(3)
1.1.2 | 2x --1 =V4— 5x ,
2 , .
(2x — 1)? = (V4—5x) squaring both sides
4x? —4x4+1+5x-4=0 Y standard form
4x? +x—-3=0 \. /
_4+./(@)2 2464) 3)
(4x -3)(x+1)=0 or x= Y factors
= cs or x#-—-1 ¥ answers with selection
‘ (4)
[7]
Question 2
P v7
‘ d)
¥ a=5 andd=-4
¥ -87 =5+(n—1)(-4)
v¥n=24
OR
v¥—-4n+ 9
“eens ¥ —4n+9=-87
an 96 Vn =A
n=
(3)
2.1.3 | -3 5-75... 5-87
n
Sn = 51a tT] Vn=2
22 faz3
S22 = = [-3 -— 87
a v answer

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OR
= 5 [2a + (n—1)d]
ae geen Yn=22
22 =
Soo =F [2(-3) + (22 - 1)-4)] Y answer
= —990
OR
Va=-3
All negative terms can be writte get | Vv Vanswer
the answer of -990
OR
vi 28
Sum = Sy4- (5+1) 2 bea
24 v6
=7[5-87]-6
= -—990 v answer
(3)
2.2.1|a+9d=21 Yat+9d=21
a+16d = 49 Yat+16d =49
“« —7d = —28
d=4 Y value of d
(3)
2.2.2} a+9(4) =21
a=-15 Ya=-15
Ya=-15
Y answer
(3)

2:34

Y all three terms

Q)

2:3:2

Y substitution and = 1189

re Y standard form
Y Factors
(2m + 41)(m — 29) =0
41

«m= 29 or m# “> Y answer
(4)
[18]

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Question 3

3.11

=58; —48

Y Both answers

d)

3,2

Ty = an? + bn+c

2a=2; 3a+b=2; at+b+c=-78

a=1; 3(1) +b » 1-1+c=-78
b=-1 c=-78

“Ty =n? —n—78

Yva=1

V¥b=-1 “c=-78
Y answer

(4)
3:2.1 || 5: 10:20%....%; =a?
Tn = (5)(2)""* Y answer
dd)
322.2 a(r™ — 1)
rns a
5[(2)8 - 1] Y substitution into the
Sie=— Jy correct formula
Sig = 1310715 v answer
(2. 4) (2x — 4) a
3.3 «i ” ae
r= a 4 Vr=2x+4
Converge: -1<r<1 OLS SY
1 <i 2a 1 Y substitution
—5 <2x < —3
5. 22 Y answer
ge" "9 (4)
[12]
Question 4
4.1 | 2x? +4++16=0
x —-2x-8=0
(x-4)(x+2)=0 Y factors
Vx=-2Vx=4
(3)
v1
v 18
(2)
vy< 8
OR
¥ ye(—0;18]

()

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44 |TP(1; 18) forf
TP(2 ; 15) forh

¥ TP (2:15) forh

6] ms v p=-l
“p=-l q=33 P
¥q=-3
(3)
AS |y=2x+4 ¥ swop x andy
x=2y+4 .
1 =-x-—
“y=sx-2 wy 2* 2
2 (2)
4.6 | g(x) =0 or g(x) =0
x=4 or x=-2 (product 0 at x vx=4
¥x=2
(2)
[13]
Question 5
5.1 |y>0 v answer
OR OR
ye(0;%) v answer
Q)
5.2
Agar
pix = 3? vx=3
= log3x
y Bs Y equation
(2)
5.3
y= log; x Y correct subst into correct
2= logs a formula (a ; 2)
Y answer
a=3*=9

(2)

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Forf
v shape
v y-intercept & a point
Y axis of symmetry/ y=x
For f-!
v shape
Y x-intercept & a point
(5)
Yes. The vertical line test cuts f~ = Y yes (2)
Y valid reason
OR OR
Yes. For every x — value there is a unique y — value v yes
Y valid reason (2)
OR OR
Yes. fis a one-to-one function v yes
¥ valid reason (2)
[12]
Question 6
6.11] y2 = /@3%= C5? [Pythagoras] yo
y =-12 y=-l2
Q
3
CS;y¥)
sin*@ Y substitution
_(_12)?
7 (- za) v answer (3)
_ 144
~ 169
OR
sin?@ = 1-—cos?6 Y square identity

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5\? Y substitution

sin?e = 1-(-=)
13

a} 144 v answer
sin*@ = Teo (3)
6.1.2 | cos(@ — 135°)
= cos0cos135° + sin@sin135° ¥ expansion

= cos@(—cos45°) + sinO(sin45°)

=Cal(-Z)+C9)

Y reduction

Y substitution
_ Ww
26 Y answer
OR
-a-a+Ca@)
13/\" ya. 13/ \2
7:
13V2 (4)
6.2 | LHS = 2cos?(45° + x)
= 2cos?(45°+x)+1-1 v+1-1
= cos[2(45°+x)]+1 Y double angle
= cos(90° + 2x) +1
= (-sin2x) +1 Y simplify
=1-sin2x Y reduction
RHS
(4)
Y compound expansion
Y subst special angle
Y simplification
v square identity
(4)

6.3.1

LHS = sin(A — B) — sin(A + B)
= sinAcosB — cosAsinB — (sinAcosB + cosAsinB)
= sinAcosB — cosAsinB — sinAcosB — cosAsinB

v sinAcosB — cosAsinB
v¥—sinAcosB — cosAsinB

= —2cosAsinB
= RHS (2)
6.3.2 | sin4dx — sin10x
= sin(7x — 3x) — sin(7x + 3x) ¥ 7x — 3x & 7x + 3x
= —2cos7xsin3x v answer
(2)

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6.3.3.| sin4x — sin 10x = sin3x
=2 cos 7x sin 3x = sin3x Y substitution

2cos 7x sin 3x + sin3x =0
sin 3x (2cos7x +1) =0

Y factorisation

x = 210° + k.360°
Or x = 330° + k.360°
x= -150° or x=-30°

x = 90° + k.360°

or x=90°

cos2x = —sinx
cos2x = cos (90° + x)
2x = 90°+x+k.360° or
2x = 360° — (90° + x) + k. 360°
x =90°+k.360° or 3x = 270° +k.360°
x = 90° + k.120°
or x=90°

x=-150° or x=-—30°

1
sin3x=0 or cos7x= 3 Y both equations
3x = 0° 7x =120° or 7x = 240° v answer
x= 0° x=17,14° x = 34,29°(N/A) v answer
(5)
[20]
Question 7
7.1.1 | 180° v answer
d)
Tl2,| 1 Y answer
Q)
7.2 |1-2sin*x =—sinx
2sin?x — sinx -1=0 Y identity
(2sinx + 1)(sinx — 1) =0 Y factors
sinx = —3 or sinx =1 Y sinx = —+
Ref 2 = 30° ref 2 = 90° Y sinx =1

¥—150° and —30° (6)
vy 90°

Y co-function

v 2x in quadrant 2

Y 2x in quadrant 3

Y both general solutions
¥—150° and —30°

¥ 90° (6)

Y co-function

¥ 2x in quadrant |

¥ 2x in quadrant 4

Y both general solutions
¥—150° and —30°

¥ 90° (6)

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OR

cos2x = —sinx
sin (90° — 2x) = —sinx
90° — 2x = 180° + x + k.360° or
90° — 2x = 360° —x + k.360°

Y co-function

¥ 90° — 2x in quadrant 3
¥ 90° — 2x in quadrant 4
Y both general solutions
¥—150° and —30°

Copyright reserved

= —30°+k.120° or x = —270° + k.360°
=-150° or x= -30° or x=90° v 90°
(6)
7.3 | A(—150°;0,5) B(—30°; 0.5)
AB = —30° — (—150°) VY AB = —30° — (150°)
AB = 120° v answer
(2)
[10]
Question 8
B11 [AC 30°
20. °°s Y trig ratio
AC = 20 cos 30°
AC = 10V3 = 17,32 units ¥ answer
(2)
OR
AC 20 Y trig ratio
sin60° sin 90°
«. AC = 20sin 60° = 17,32 units Y answer
(2)
8.1.2 | AB? = AC? + BC? — 2AC. BCcos ACB Y cosine formula
AB2 = (10v3)° + (8) — 2(10V3)(8)Ccos 100° Y substitution into cosine
. formula
AB = 20,30 units v
answer
(3)
8.2 |sinADB sin ABD
AB. AD Y sine formula in AABD
sinADB_ sin73,4° Y substitution in the sine
—__ ra formula
v answer
(3)
[8]
TOTA L: 100

10

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